[poca]

50ml .1M HNO2
5ml .1M NaOH

To get pH

pH = -log(7.1*10^-4) + log [ .0091 / .00818]
pH = 3.2

[poca]

I'm also getting 2.18 but the book answer is 2.38, which I can't get

[Borek]

Book is right.

.0091 / .00818

Explain where did you get these numbers from.

[poca]

HA = [ (50*.1) - (5*.1) ] / 55 = .00818
A- = [ 5*.1] / 55 = .0091

[gertrudetrumpet]

Are you sure that the concentration of ha and a- are those values? What is the relationship between the two species?

[poca]

The Ka would be a relationship. so are you saying

7.1*10^-4 =( [H+][A-] ) / (.00818)  with [H+] = [A-] ?

maybe [A-] = .0024?

[gertrudetrumpet]

If hno2 is ha, what is a-?

[poca]

NO2 -
so the overall equation is..

HNO2 + NaOH = H2O + NaNO2

[Borek]

HA = [ (50*.1) - (5*.1) ] / 55 = .00818
A- = [ 5*.1] / 55 = .0091

Your approach is correct, unfortunately, acid is too strong for the approximation you are using. Concentration of A^- is higher than expected, as acid is not only neutralized, but also further dissociates on its own.

[gertrudetrumpet]

right, so after it neutralizes, it will also go back to hno2, that is when you use the ka value and calculate for the final equilibrium concentrations of no2 and hno2, or a- and ha for the hh equation.